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Then Therefore, M j mj sup sup Mj D D j ˚ j ˚ x; x0 xj x; x0 xj f .x0/ j f .x0/ j xj 1 1 xj ˇ ˇ ˇ ˇ (3.3.4) S .P / s.P / S.P / s.P /; and those on the right are where the upper and lower sums on the left are associated with associated with f . Now suppose that > 0. Since f is integrable on Œa; b, Theorem 3.2.7 s.P / <... |
x/ D a Z (3.3.18) g.x/; a < x < b (Theorem 3.3.11). Therefore, Theorem 3.3.15 with u then G0.x/ and v D G yields D f D b a Z f .x/g.x/ dx D b b a a Z f 0.x/G.x/ dx: (3.3.19) Since f 0 is nonnegative and G is continuous, Theorem 3.3.7 implies that f .x/G.x.x/G.x/ dx D b G.c/ f 0.x/ dx a Z (3.3.20) 146 Chapter 3 Integral... |
ulate Definition 3.1.1 for a function defined on an infinite or semiinfinite interval would introduce questions concerning convergence of the resulting Riemann sums, which would be infinite series. In this section we extend the definition of integral to include cases where f is unbounded or the interval is unbounded, or both... |
4.4) Theorem 3.4.6 and the first inequality in (3.4.4) imply that b a2 Z g.x/ dx < if 1 b a2 Z f .x/ dx < : 1 160 Chapter 3 IntegralCalculusofFunctionsofOneVariable Theorem 3.4.6 and the second inequality in (3.4.4) imply that b a2 Z f .x/ dx < if 1 b a2 Z g.x/ dx < : 1 Therefore, they must diverge to f .x/ dx and b a2 ... |
one on the left. Example 3.4.17 The integral converges if and only if p > 1 (Example 3.4.3). Defining .t/ Theorem 3.4.13 yields D 1=t and applying 1 x p dx 1 Z which implies that 1 x p dx D 1 t p 2 t dt dt converges if and only if q > R 1 t p 2 dt; 0 Z 1. 3.4 Exercises 1. (a) Let f be locally integrable and bounded on Œ... |
ntinuity from the right or left, and only if wf .x0/ respectively./ D Proof Suppose that a < x0 < b. First, suppose that wf .x0/ 0 and > 0. Then D for some h > 0, so Wf Œx0 h; x0 C h < f .x/ f .x0/ j x0, we conclude that j Letting x0 D < if x0 h x; x0 x0 C h: Therefore, f is continuous at x0. j f .x/ f .x0/ j < if x j ... |
domain. ˙1 SECTION 4.3 introduces concepts of convergence and divergence to for infinite series of constants. We prove Cauchy’s convergence criterion for a series of constants. In connection with series of positive terms, we consider the comparison test, the integral test, the ratio test, and Raabe’s test. For general ... |
hat t =2 =2; thus, there is an integer j j tn t j j j j tn j t j .tn t/ t j j 2 t tn t if n M: j tn j j j D j j jj < if n C If > 0, choose N0 so that 1 tn ˇ ˇ 1=t. Now we obtain (4.1.9) in the general case from (4.1.8) with ˇ ˇ jj N0, and let N max.N0; M /. Then t tn j j 2 2 t 1 t if n 1=tn tn hence, limn tn !1 replace... |
positive numbers. Let D 0/, then 1xn .n D C 1 g tan y such that xn converges. (d) sn nŠ nn .2n/Š 22n.nŠ/2 D D =2 < y < =2. Prove: sn 1 C 1 2 D sn ; A sn C 0: n 0. pA if n sn if n 1. sn exists. 1 limn !1 1 C D (a) Show that sn C (b) Show that sn (c) Show that s (d) Find s. sn Prove: If . 1 Prove Theorem 4.1.7. f g is u... |
teger N such that f .x/ imply that j j condition is necessary. < if n f .xn/ f .x/ j f .x/ j < if j x < ı: x j (4.2.8) N . This and (4.2.8) N . This implies (4.2.6), which shows that the stated < ı if n xn j j x For sufficiency, suppose that f is discontinuous at x. Then there is an 0 > 0 such that, for each positive in... |
for convergence; an an an cannot converge unless limn 0. The condition is not sufficient; that is, may diverge even if limn P an D !1 D 0. We will see examples below. !1 We leave the proof of the following corollary of Theorem 4.3.5 to you (Exercise 4.3.5). P Corollary 4.3.7 If that P that is; an converges; then for ea... |
The ratio test does not imply that an D 1 converges if 0 < r < 1 or diverges if nr n 1 n C r; .n P 1/r n C nr n 1 r: 1 D an 1 C an D lim n !1 D 1, but then Corollary 4.3.6 implies that the series if merely an < an C an 1 1 < 1 P (4.3.14) 1, in which case the test is if (4.3.14) is replaced by 1 for large n, since this... |
3 Suppose that an increasing sequence of integers, with n1 1n D k an A; where D k. Define 1 A : Let 1 nj g11 be f an1; C C 1 C C C an2 ; P ak an1 b1 b2 D D ::: Then br anr 1 1 C D C C anr : bnj D A: 1 1 j X D Proof If Tr is the r th partial sum of 1 bnj and 1j D An f g is the nth partial sum of k as, then 1s D P Tr D D ... |
due to Cauchy. We will see the P importance of this definition in Section 4.5 Definition 4.3.28 The Cauchy product of 0 an and 1n D cn D a0bn C a1bn 1 P C C Thus, cn is the sum of all products ai bj , where i 1b1 an 0, j n 1n D anb0: P C 0, and i 0 bn is 0 cn, where 1n D P (4.3.38) j C D n; thus, (4.3.39) cn D n 0 r X D... |
0 n X D converges to AB. HINT: Let Show that f An Cn AnB D and apply Theorem 4.3.5 to an . j j P n 0 r X D ar .Bn r B/ 234 Chapter 4 InfiniteSequences andSeries n 1 ar 0 r;s X D a.i / j D 1j D 41. Suppose that ar lim n !1 1 n 0 for all r 0 and and 10 ar A < . Show that 1 0 and s C D P lim n !1 1 n ar s D 2A a0: D n 1 0... |
Fn f g Proof Suppose that each Fn is continuous at x0. If x S and n F .x/ j F .x0/ j j j F .x/ Fn.x/ Fn.x/ Fn.x0/ Fn.x/ Fn 2 k j C j j C Suppose that > 0. Since Fn g S < . For this fixed n, (4.4.8) implies that F f k k Fn Fn.x0/ 1, then F .x0/ j (4.4.8) j C j 2 Fn.x0/ S : F k converges uniformly to F on S , we can choos... |
j .x Xj D fj .x/ fj 1.x/ j C ; S; x 2 (4.4.21) (4.4.22) (4.4.23) Section 4.4 Sequences andSeriesofFunctions 249 so m m .fj fj M fj fj / converges absolutely uniformly on S , TheNow suppose that > 0. Since C orem 4.4.13 implies that there is an integer N such that the right side of the last inequality is less than if m ... |
es uniformly on S and S , then gn fn S k k k k (b) This “comparison test” can be corrected by adding one word to its hypothesis and conclusion. What is the word? (a) Explain the difference between the following statements: (i) fn converges fn converges absolutely uniformly absolutely and uniformly on S ; (ii) on S . P ... |
he induction. C Example 4.5.6 In Example 4.4.10 we saw that 1 1 x D 1 xn; 0 n X D < 1: x j j Repeated differentiation yields kŠ x/k 1 D C .1 D 1 k Xn D 1 0 n X D n.n 1/ .n k C 1/xn k k/.n .n C k C 1/ .n C 1/xn; < 1; x j j 1 x/k . ! xn; < 1: x j j so Example 4.5.7 By the method of Example 4.5.5, it can be shown that the... |
equations successively yields a0 a1 a2 a3 D D D D so 1 4 ; 1 2 a0 2 D 1 a0 2 C 2 a0 1 6 C 2 ; a1 1 2 D a2 D 1 4 1 2 1 4 0; D 1 12 1 8 C 0 D 1 48 ; a1 2 C 1 2 x 4 C x3 48 C : ex D 1 1 C 272 Chapter 4 InfiniteSequences andSeries Example 4.5.15 To find the reciprocal of we again let h D 1 in (4.5.25). If g.x/ ex D D xn nŠ ;... |
he sum of two vectors, the product of a vector and a real number, the length of a vector, and the inner product of two vectors. We study the arithmetic properties of Rn, including Schwarz’s inequality and the triangle inequality. We define neighborhoods and open sets in Rn, define convergence of a sequence of points in R... |
e the parametric equation of a line through two points X0 and X1 in R3, we take U D X1 0 in (5.1.9), which yields X D X0 C t.X1 X0/ tX1 .1 C D t/X0; < t < : 1 1 The line segment from X0 to X1 consists of those points for which 0 t 1. Example 5.1.5 The line L defined by x 1 D C which can be rewritten as 2t; y 3 D 4t; ´ 1... |
for this. If we wish to raise questions of continuity and differentiability at every point of the domain D of a function f , then every point of D must be a limit point b of D0. Intervals have this property. Moreover, the definition of a f .x/ dx is obviously applicable only if f is defined on Œa; b. It is not productive... |
.1 Structureof Rn 301 23. 22. Use the Bolzano–Weierstrass theorem to show that if S1, S2, . . . , Sm, . . . is an , then 1 Sm is nonempty. Show that the conclusion does not follow if the sets are infinite sequence of nonempty compact sets and S1 Sm S2 1m D assumed to be closed rather than compact. T Suppose that a seque... |
1 a2 C a 1 : Section 5.2 ContinuousReal-Valued Functionsof n Variables 309 y y = ax x Figure 5.2.2 We leave it to you to define lim j cise 5.2.6). f .X/ X j!1 D 1 and lim j X j!1 f .X/ D 1 (Exer- We will continue the convention adopted in Section 2.1: “limX X0 f .X/ exists” means , we D X0 f .X/ exists in the extended ... |
ned so as to be continuous. 10. Repeat Exercise 5.2.9 for the functions in Exercise 5.2.5. 11. Give an example of a function f on R2 such that f is not continuous at .0; 0/, / and f .x; 0/ is a continuous but f .0; y/ is a continuous function of y on . ; function of x on . 1 1 /. ; 1 1 Prove Theorem 5.2.11. HINT: See t... |
hen and xy f .x; y/ y2 x2 0; C D ( ; .x; y/ .x; y/ .0; 0/; .0; 0/; ¤ D (5.3.15) fx.0; 0/ D lim 0 h ! fy.0; 0/ D lim 0 k ! f .h; 0/ f .0; k/ h k f .0; 0/ 0 0 h D 0 D lim 0 h ! f .0; 0/ 0 0 k D 0; D lim 0 k ! but f is not continous at .0; 0/. (See Examples 5.2.3 and 5.2.11.) Therefore, if differentiability of a function ... |
y/2 y2 C D 8 < 0; p : sin x 1 y ; x x y; y; ¤ D .x x2 y/2 y2 C Therefore, ˇ ˇ ˇ ˇ ˇ p sin x 1 2.x2 x2 y2/ y2 x2 C y2; x y: ¤ p f .x; y/ f .0; 0/ .x;y/ .0;0/ lim ! fx.0; 0/x y2 x2 C fy.0; 0/y 0; D so f is differentiable at .0; 0/, but fx and fy are not continuous at .0; 0/. p Geometric Interpretation of Differentiabilit... |
n the proof of Theorem 5.3.3 to write C C C C B.h/ D Œfx. x; y0 k/ C fx. x; y0/ h; x is between x0 and x0 where .x0; y0/ to infer that b C h. Then use the differentiability of fx at b b B.h/ D h2fxy.x0; y0/ C hE1.h/; where E1.h/ h D 0: lim 0 h ! (b) Use the mean value theorem to write B.h/ D fy.x0 y is between y0 and y... |
2 2 @x @r D cos ; @y @r D sin ; and @2f @x @y D @2f @y @x S (Exercise 5.3.21), (5.4.19) yields @2h @r 2 D cos2 @2f @x2 C 2 sin cos @2f @x @y C sin2 @2f @y2 : Differentiating (5.4.11) with respect to yields cos cos @2h @ @r D sin sin D sin C @f @x C @f @x C @2f @x @y @f @y C @f @y C @2f @y2 @x @ C cos cos @y @ : @ @ @f... |
te. The polynomial is not semidefinite, since, for example, p2.x; y; ´/ x2 y2 C ´2 D p2.1; 0; 0/ D 1 and p2.0; 1; 0/ 1: D From Theorem 5.3.11, if f is differentiable and attains a local extreme value at X0, then dX0f 0; D (5.4.37) since fx1.X0/ 0. However, the converse is false. The next D theorem provides a method for ... |
s. The differential of a vector-valued function F is defined as a certain linear transformation. The matrix of this linear transformation is called the differential matrix of F, denoted by F0. The chain rule is extended to compositions of differentiable vector-valued functions. SECTION 6.3 presents a complete proof of t... |
ame number of rows and columns. We assume that you know the definition of the determinant det.A/ of an n n matrix D a11 a12 a21 a22 ::: ::: an2 an1 a1n a2n ::: ann :: : ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ a12 a22 ::: an2 : :: a1n a2n ::: ann a11 a21 ::: an1 A D 2 6 6 6 4 370 Chapter 6 Vector-Valued FunctionsofSeveral Variables The transp... |
B.X; Y/ D A 1.X/A.Y/ I: Show that for each > 0 there is a ı > 0 such that if X; Y K and X j Y j 2 bij .X; Y/ j < ; 1 i; j m; j < ı. HINT: Show that bij is continuous on the set .X; Y/ X K; Y K : 2 2 Then assume that the conclusion is false and use Exercise 5.1.32 to obtain a contradiction: ˚ ˇ ˇ 6.2 CONTINUITY AND DIFF... |
; X: Y D Y. We now show that if X0 D, 2 .X;Y/ .X0;X0 / lim ! g.X; Y/ 0 g.X0; X0/ I D D (6.2.19) that is, g is also continuous at points .X0; X0/ in S . Suppose that > 0 and X0 D. Since the partial derivatives of f1, f2, . . . , fm are 2 continuous on an open set containing D, there is a ı > 0 such that @fi .Y/ @xj @fi ... |
x; y/ @x @f2.x; y/ @x D D D D 2x; uy .x; y/ @f1.x; y/ @y D 2y; D 2x; vy.x; y/ @f2.x; y/ @y D 2y: D To find F.R2/, we observe that so u v C D 2x2; u v D 2y2; F.R2/ T .u; v/ u v 0; u v which is the part of the uv-plane shaded in Figure 6.3.1. If .u; v/ D C ˚ 0 ; T , then 2 pu C 2 v ; F so F.R2/ T . D ˇ ˇ pu Figure 6.3.1 3... |
A is nonsingular, in which case R.F/ Rn and D F 1 are the differential matrices of F and F A D 1.U/ 1U: 1, respectively, we can say that a Since A and A linear transformation is invertible if and only if its differential matrix F0 is nonsingular, in 1 is given by which case the differential matrix of F .F 1/0 D .F0/ 1:... |
ere k is a nonzero integer. No subset containing the origin .x; y/ .0; 0/ has this property, nor does any deleted neighborhood of the origin (Exercise 6.3.14), so there are open sets on which no branch of the argument can be defined. However, if one branch can be defined on T , then so can infinitely many others. (Why?) A... |
s of X by x, y, . . . , and the components of U by u, v, . . . . To motivate the problem we are interested in, we first ask whether the linear system of m equations in m n variables C a12x2 a22x2 a11x1 a21x1 C C a1nxn a2nxn C C C C am1x1 am2x2 C C C amnxn b11u1 b21u1 bm1u1 C C C C C C b12u2 b22ux C C C C b1mum b2mum bm2... |
x; y; u.x; y// fu.x; y; u.x; y// D ; uy.x; y/ fy.x; y; u.x; y// fu.x; y; u.x; y// : D In particular, since u.1; 1/ D 1, ux.1; 1/ D Example 6.4.3 Let X D 2 4 and 0 7 D x y ´ 3 5 2x2 0; uy.1; 1/ D 4 7 D 4 7 : and U u v ; D F.X; U/ D y2 ´2 u2 C ´2 C 2u v C x2 v2 : C .0; 2/, then F.X0; U0/ C 0. Moreover, D If X0 .1; D 1; 1... |
; y; ´; u; v/ 0; y.x0; v0/ u0; ´.x0; v0/ ´0; u.x0; v0/ u0 D D D D determine y, ´, and u as continuously differentiable functions of .x; v/ near .x0; v0/. Use Cramer’s rule to express their first partial derivatives as ratios of Jacobians. 14. Decide which pairs of the variables x, y, ´, u, and v are determined as functi... |
0 of f over P such that R1; R2; : : : ; Rk D f g is any parti- This implies that f cannot satisfy Definition 7.1.2. (Why?) j 0 j M: (7.1.11) Let D k f .Xj /V .Rj / 1 j X D be a Riemann sum of f over P . There must be an integer i in f .X/ j f .Xi / j M V .Ri / 1; 2; : : : ; k such that g f (7.1.12) for some X in Ri , be... |
R02, . . . , R0p with V .R0j / < 2p ; 1 p: j p (7.1.27) Now, m E 1 i [ D Ri ! [ and, from (7.1.26) and (7.1.27), 0 1 j [ D @ R0j 1 A V .Ri / R0j / < : Example 7.1.5 If f is continuous on Œa; b, then the curve f .x/; a y D x b /, has zero content in R2. To see this, (7.1.28) y .x; y/ b (that is, the set suppose that > 0... |
roof. D ZT Theorem 7.1.30 If f is integrable on disjoint sets S1 and S2; then f is integrable on S1 S2; and [ Proof For i D S2 [ ZS1 1, 2, let f .X/ d X D ZS1 f .X/ d X C ZS2 f .X/ d X: (7.1.39) fSi .X/ f .X/; X D ( 0; X Si ; Si : 2 62 From Lemma 7.1.29 with S Si and T S1 [ D D S2, fSi is integrable on S1 S2, and [ fSi... |
rtition P of R such that Sf .P/ sf .P/ < , from Theorem 7.1.12. Consequently, from (7.2.6), there is a partition P2 of Œc; d such that SF .P2/ sF .P2/ < , so F is integrable on Œc; d , from Theorem 3.2.7. It remains to verify (7.2.1). From (7.2.4) and the definition of d c F .y/ dy, there is for R each > 0 a ı > 0 such ... |
1 .y3 0 Z y5/ dy xy d.x; y/ ZS D D y 1 y x2 2 0 ˇ ˇ y4 ˇ ˇ 4 Z 1 2 y2! y6 6 1 0 D 1 24 : 1, 1) S x Figure 7.2.3 In this case we can also represent S in the form (7.2.21) as S D .x; y/ x y px; 0 x 1 hence, from (7.2.22), ˚ ˇ ˇ I xy d.x; y/ ZS D D 1 px x dx y dy 1 .x2 x Z x3/ dx Example 7.2.7 To evaluate 1 y2 2 0 x3 3 y ... |
4x2 y2 4y2 y2 and ´ x2 y2; .x; y; 0/ is in the triangle with vertices y2. x2 C D 8 ˇ ˇ ˇ ˇ ˇ ˇ 484 Chapter 7 IntegralsofFunctionsofSeveral Variables 19. Let R Œa1; b2 x2 D R .x1 C (a) Œa2; b2 xn/ d X (c) R R x1x2; C C xn d X 20. Assuming that f is continuous, express R Œan; bn. Evaluate R .x2 (b) R 1 C x2 2 C C x2 n/ d... |
j // V .L.Rj //: (7.3.18) U1 Xj 2 U2 Xj 2 If we assume that (7.3.15) holds whenever R is a rectangle, then U1 Xj 2 C V .L.Rj // det.A/ j D j V .Rj /; 1 j k; so (7.3.18) implies that This, (7.3.16) and (7.3.17) imply that s.P / V .L.S // det.A/ j j S.P /: hence, since can be made arbitrarily small, (7.3.14) follows for ... |
f .X/ d X ZS f .G.Y// j J G.Y/ j d Y: (7.3.37) Then Q.S / 0: Proof From the continuity of J G and f on the compact sets S and G.S /, there are constants M1 and M2 such that and J G.Y/ j j M1 if Y S 2 (7.3.38) 2 (Theorem 5.2.11). Now suppose that > 0. Since f (Theorem 5.2.14), there is a ı > 0 such that j j f .X/ M2 if ... |
.T / D d X ZG.S / a r 2 dr D 0 Z D ZS =2 r 2 cos d.r; ; / =2 d cos d 0 Z 0 Z a3 3 2 D (7.3.1) a3 6 : D Example 7.3.5 Evaluate the iterated integral I D 0 Z a x dx pa2 x2 0 Z dy 0 Z pa2 x2 y2 ´ d´ .a > 0/: Solution We first rewrite I as a multiple integral where G and S are as in Example 7.3.4. From Theorem 7.3.15, I D Z... |
ion to vector spaces over the real numbers. simply as R.) D v j j U; C C C V V .V 2 D C Definition 8.1.2 A vector space A is a nonempty set of elements called vectors on which two operations, vector addition and scalar multiplication (multiplication by real numbers) are defined, such that the following assertions are tru... |
S . The set of boundary points of S is the boundary of S , denoted by @S . The closure of S , denoted by S , is defined by S S S and there is a neighborhood of u0 that contains @S . D [ (c) u0 is an isolated point of S if u0 no other point of S . 2 (d) u0 is exterior to S if u0 is in the interior of S c. The collection ... |
/ D 0 if and only if u .w; u/ D .w; v/. C v; 2. Prove: If x, y, u, and v are arbitrary members of a metric space .A; /, then .x; y/ j .u; v/ j .x; u/ C .v; y/: 3. (a) Suppose that .A; / is a metric space, and define 1.u; v/ .u; v/ D 1 C .u; v/ : Show that .A; 1/ is a metric space. (b) Show that infinitely many metrics ca... |
f and only if every infinite sequence of members of T has a subsequence that converges to a member of T: tn f g Proof Suppose that T is compact and terms, there is a t in T such that tn n1 < n2 < tnj D has a limit point t in T , so there are integers n1 < n2 < therefore, limj has only finitely many distinct t for infinite... |
are nonempty subsets of a metric space .A; /, we define the distance from S to T by ˇ ˇ ˚ dist.S; T / D inf .s; t/ s S; t 2 Show that if S and T are compact, then dist.S; T / some t in T . ˇ ˇ ˚ D 4. (a) Show that every totally bounded set is bounded. (b) Let ıi r 1 0 D if i if i r; r; D ¤ T : 2 .s; t/ for some s in S a... |
no) (f ) p7 (no); (c) p7 (yes); p7 (no) p7 (yes) Section 1.2 pp. 15–19 1:2:9 (p. 16) (a) 2n=.2n/Š (b) 2 3n=.2n 1/Š C (c) 2 n.2n/Š=.nŠ/2 (d) nn=nŠ 1:2:10 (p. 16) (b) no 1:2:11 (p. 16) (b) no 1:2:20 (p. 18) An xn nŠ 0 ln x D 1:2:21 (p. 18) fn.x1; x2; : : : ; xn/ 1 min.x1; x2; : : : ; xn/ 2n @ max.x1; x2; : : : ; xn/, gn.... |
in.x=n/, Gn.x/ FnGn k 1 . D 1 S D k (d) e 1 2 ; / 1 / (e) Œr; (b) Œ 1 2 ; / (c) closed sub- 1 /; r > 1 (f ) compact subsets 1 an (constant), where an converges fn.x/ con- P converges pointwise and fn.x/ j 1 converges uniformly on S . x2n C 1/.2n . 1/n .2n 1/b) P 1 C x2n nŠ.2n 1/ C 556 AnswerstoSelected Exercises Sectio... |
4 y:4:3 (p. 431) fi .X; U sin x sin y u v x y (c) r aij .xj xj 0/ 1 .ui 0. (a ui 0/s, 1 i m, where r 3; (b) r 1, s D D 3; (c and s are positive integers and not all aij r 6:4:4 (p. 431) ux.1; 1/ 6:4:5 (p. 431) ux.1; 1; 1/ 6:4:6 (p. 431) (a) u.1; 2/ (b) u. 1; 2/ 1; D D (c) u.=2; =2/ ux.=2; =2/ (d) u.1; 1/ 5 8 , uy.1; 1... |
rcise 5.1.25) between two vectors, 283 Distributive law, 2 (see p. 1) Divergence, unconditional, 233 (Exercise 4.3.38) Index 569 E Edge lengths of a coordinate rectangle, 437 Elementary matrix, 488 Empty set, 4 Entries of a matrix, 364 -neighborhood, 21, 289, 525 -net, 539 Equicontinuous subset of C Œa; b, 541 Equivale... |
rocal, 271 of a quotient, 269 uniqueness of, 263 Prime, 15 Principal value, 155 Principle of mathematical induction, 11, 14 Principle of nested sets, 530 Product Cartesian, 31, 436 (see p. 435) Cauchy, 226, 233 (Example 4.3.40) inner, 284 of matrices, 364 of power series, 268 of series, 223 Proper integral, 153 R Rn, 2... |
uch that for each pair x,y of vertices of G, xy ∈ E(G) if and only if f (x)f (y) ∈ E(H). In other words, G and H are isomorphic if there exists a mapping from one vertex set to another that preserves adjacencies. The mapping itself is called an isomorphism. In our example, such an isomorphism could be described as foll... |
owing graphs: P2k, P2k+1, C2k, C2k+1, Kn, Km,n. 3. For each graph in Exercise 2, find the number of vertices in the center. 4. If x is in the periphery of G and d(x, y) = ecc(x), then prove that y is in the periphery of G. 1.2 Distance in Graphs 21 5. If u and v are adjacent vertices in a graph, prove that their eccentr... |
Models and Distance Do I know you? — Kevin Bacon, in Flatliners We have already seen that graphs can serve as models for all sorts of situations. In this section we will discuss several models in which the idea of distance is significant. The Acquaintance Graph “Wow, what a small world!” This familiar expression often f... |
ices should be relatively small. The characteristic path length of a graph G, denoted LG, is the average distance between vertices, where the average is taken over all pairs of distinct vertices. In any graph of order n, there are |E(Kn)| distinct pairs of vertices, and in Exercise 1 of Section 1.1.3, you showed that |... |
mponent, implying that G is a tree. It is not uncommon to look out a window and see leafless trees. In graph theory, though, leafless trees are rare indeed. In fact, the stump (K1) is the only such tree, and every other tree has at least two leaves. Take note of the proof technique of the following theorem. It is a stand... |
of minimum total weight. Proof. Let G be a connected, weighted graph of order n, and let T be a spanning tree obtained by applying Kruskal’s algorithm to G. As we have seen, Kruskal’s algorithm builds spanning trees by adding one edge at a time until a tree is formed. Let us say that the edges added for T were (in ord... |
ed, we know that k is at least n − 1. Let N be the n × k matrix whose (i, j) entry is defined by [N ]i,j = 1 if vi and fj are incident, 0 otherwise. N is called the incidence matrix of G. Since every edge of G is incident with exactly two vertices of G, each column of N contains two 1’s and n − 2 zeros. Let M be the n ×... |
umber of times. Before I turn to the problem of finding such a sequence, it would be useful to find out whether or not it is even possible to arrange the letters in this way, for if it were possible to show that there is no such arrangement, then any work directed towards finding it would be wasted. I have therefore tried... |
, by patching the circuit Qi into Ri at vi. vi. Increment i by 1, and go to step ii. An example of this process is shown in Figure 1.55. You should note that the 58 1. Graph Theory @ @ @ @ @ @ @ FIGURE 1.55. The stages of Hierholzer’s algorithm. process will succeed no matter what the initial circuit, R1, is chosen to ... |
ibed, then κ(G) ≥ 2 — for if κ(G) = 1, then α(G) = 1 and thus G is either K1 or K2, contradicting the fact that n ≥ 3. Let C be a longest cycle in G. Suppose that C is not a Hamiltonian cycle, and let v be a vertex of G that is not on C. Let H be the connected component of 64 1. Graph Theory G − V (C) that contains v. ... |
the respective authors showed that the pairs {K1,3, W }, {K1,3, P6}, and {K1,3, Z2} (see Figure 1.69) all imply Hamiltonicity when forbidden in 2-connected graphs. Do you see a pat- . - FIGURE 1.69. Two additional forbidden subgraphs. tern here? The claw seems to be prominent in results like this. In 1997, Faudree and ... |
of R, denoted by b(R), to be the number of edges that bound region R. For example, in Figure 1.75, b(R1) = b(R4) = 4, b(R2) = b(R3) = b(R5) = b(R6) = 3, and b(R7) = 12. In Figure 1.76, b(S1) = b(S3) = 3 and b(S2) = 6. Note that in this graph, the edges e1, e2, and e3 do not contribute to the bound degree of any region... |
e obtain rV = 2E = kF . Now, Theorem 1.36 implies that 8 = 4V − 4E + 4F = 4V − 2E + 4F − 2E = 4V − rV + 4F − kF = (4 − r)V + (4 − k)F. 1.5 Planarity 83 V and E are of course both positive, and since 3 ≤ k ≤ 5 and r ≥ 3, there are only five possible cases. Case 1. Suppose r = 3 and k = 3. In this case, V = F and 8 = V + ... |
≥ k − 1. (e) Find all of the 3-critical graphs. Hint: Use part (d). 1.6.2 Bounds on Chromatic Number The point is, ladies and gentlemen, that greed, for lack of a better word, is good. Greed is right. Greed works. — Gordon Gekko, in Wall Street In general, determining the chromatic number of a graph is hard. While sma... |
in a detour path (a longest path) of G, prove that χ(G) ≤ τ (G). 5. Prove that the only graph G of order n for which χ(G) = n is Kn. 6. Prove that for any graph G of order n, n α(G) ≤ χ(G) ≤ n + 1 − α(G). 7. If G is bipartite, prove that ω(G) = χ(G). 8. Let G be a graph of order n. Prove that (a) n ≤ χ(G)χ(G); √ (b) 2... |
thought of as a coloring of G − e where u and v are colored the same. Thus, the answer to this question is cG/e(k). 1.6 Colorings 99 D E FIGURE 1.98. Examples of the operations. Now, how many k-colorings are there of G − e where u and v are assigned different colors? If C is a such a coloring of G − e, then C can be co... |
the other? The answer to this question is found in the following result of Hall [147] (Philip, not Monty). Recall that the neighborhood of a set of vertices S, denoted by N (S), is the union of the neighborhoods of the vertices of S. 106 1. Graph Theory Theorem 1.51 (Hall’s Theorem). Let G be a bipartite graph with par... |
ph of order 2n such that δ(G) ≥ n, then G has a perfect matching. Proof. Let G be a graph of order 2n with δ(G) ≥ n. Dirac’s theorem (Theorem 1.22) guarantees the existence of a Hamiltonian cycle, C. A perfect matching of G is formed by using alternate edges of C. In 1947 Tutte [269] provided perhaps the best known cha... |
llest integer n such that every 2-coloring of the edges of Kn either contains a red Kp or a blue Kq as a subgraph. Read through that definition at least one more time, and then consider this simple example. We would like to find the value of R(1, 3). According to the definition, this is the least value of n such that ever... |
ent was due in part to the search for the elusive classical Ramsey numbers, for it was thought that the more general topic might shed some light on the search. The generalization blossomed and became an exciting field in itself. In this section we explain the concept of graph Ramsey theory, and we examine several result... |
ints lie on the same line, can we arrange a large number of points so that no subset of them forms the vertices of a convex octagon? 2.1 Some Essential Problems The mere formulation of a problem is far more essential than its solution. . . — Albert Einstein 2.1 Some Essential Problems 131 We begin our study of combinat... |
racters? (c) How many are there with at most five characters, if they must read exactly the same forwards and backwards? For example, kayak and T55T are admissible, but Kayak is not. 2. Assume that a vowel is one of the five letters A, E, I, O, or U. (a) How many eleven-letter sequences from the alphabet contain exactly ... |
in by enter= 1 for n ≥ 0; then use (2.6) ing 1 in the first position of each row, since to compute the entries in successive rows of the table. The resulting pattern of numbers is called Pascal’s triangle, after Blaise Pascal, who studied many of its properties in his Trait´e du Triangle Arithm´etique, written in 1654. ... |
nomial coefficients from Section 2.2. We begin with a more general formula for expanding multinomial coefficients in terms of factorials. Expansion. If n is a nonnegative integer, and k1, . . . , km are integers satisfying k1 + · · · + km = n, then n k1, . . . , km ⎧ ⎨ = ⎩ n! k1! · · · km! 0 if each ki ≥ 0, otherwise. (2... |
ogists asserts that the number of five-letter sequences that can be formed using the letters of the Hawaiian long-nosed butterfly fish, the lauwiliwilinukunuku’oi’oi, is more than twice as large as the number of five-letter sequences that can be created using the name of the state fish of Hawaii, the painted triggerfish humu... |
teed by the theorem satisfies α − p q < 1 q2 + q . (2.25) Exercise 11 asks you to show that there exist infinitely many rational numbers p/q that satisfy this inequality for a fixed irrational number α. Exercises 1. Show that at any party with at least two people, there must exist at least two people in the group who know... |
nt the number of elements that remain √ when multiples of prime numbers p ≤ n are excluded from the set. Since every composite number m ≤ n has a prime factor p ≤ m, excluding all of these numbers removes all the composite numbers from the set. m. √ √ For example, for n = 120, the largest prime less than or equal to n ... |
l clothing. At intermission, the panelists decide to rearrange themselves so that it will be apparent to the audience that everyone has moved to a different seat when the panel reconvenes. Each twin can therefore take neither her own former place, nor her twin’s. Let Tn denote the number of different ways to derange th... |
Monty Python and the Holy Grail Suppose there is an inexhaustible supply of each of n different objects. How many ways are there to select m objects from the n different objects, if you are allowed to select each object as many times as you like? Let an,m denote this number. Evidently, for fixed n, the generating funct... |
llar in change. This is a fairly efficient method to determine ak, since apparently we can calculate this number using at most 5k arithmetic operations. But we can do much better! We can compute ak using at most a constant number of arithmetic operations, regardless of the value of k. To show this, let us first simplify ... |
− ϕ x + ϕ − 1 1 + x/ϕ − 1 1 − ˆϕx , G(x) = = = since ϕ ˆϕ = −1. Now the two terms on the right are closed forms for simple geometric series, so G(x) = 1√ 5 and therefore (ϕk − ˆϕk)xk, k≥0 ϕk − ˆϕk √ 5 . (2.41) Fk = √ Notice that | ˆϕ| < 1, so Fk ∼ ϕk/ 5: a large number of rabbits indeed. Exercises 1. In each of the fol... |
pair of parentheses is x0(x1x2x3x4), another five for (x0x1x2x3)x4, two for (x0x1)(x2x3x4), and two more for (x0x1x2)(x3x4). We record these numbers in the following table. k Ck 1 0 1 1 2 2 5 3 14 4 Can we determine a recurrence relation for Ck? Suppose we group the terms so that the last multiplication occurs between x... |
f the ith object in the ordering occupies the jth position in the permutation. For example, the permutation [c, d, a, e, b] of the list [a, b, c, d, e] is represented by the function π defined on the set {1, 2, 3, 4, 5}, with π(1) = 3, π(2) = 5, π(3) = 1, π(4) = 2, and π(5) = 4. Notice that a function π : {1, . . . , n}... |
e, suppose S = {1, 2, 3, 4} is the set of vertices of the square in Figure 2.5, and C is the set of all possible colorings of these vertices using two colors, red and green. Let rrgr denote the coloring where vertices 1, 2, and 4 are red and vertex 3 is green. Then C = {gggg, gggr, ggrg, ggrr, grgg, grgr, grrg, grrr, r... |
Four rotations, k = 2, 6, 14, and 18, make two cycles of length 10, contributing 4x2 10. Rotations with k = 4, 8, 12, or 16 make 202 2. Combinatorics four cycles of length 5, adding 4x4 4. The rotation with k = 10 yields x10 1 . Ten of the reflections, the ones about axes of symmetry that pass through midpoints of edge... |
otes a large portion of his paper [224] to applications involving enumeration of graphs, trees, and chemical isomers. Exercises 1. What is the pattern inventory for coloring n objects using the m colors y1, y2, . . . , ym if the group of symmetries is Sn? 2. Use P´olya’s enumeration formula to determine the number of s... |
e the set of colorings c for which π and ρ have the same effect, Uπ = {c ∈ C : π∗(c) = ρ∗(c)}. Note that if c ∈ Uπ then automatically ρ∗(c) ∈ c. Thus, we find that FG,ρ(y) = 1 |G| π∈G c∈Uπ yv(c). (2.65) Now suppose π ∈ G, and π has λi cycles of length i, for each i with 1 ≤ i ≤ n. Let i denote the length of the ith cycl... |
] + 25236[r12q7l] + 88620[r12q6l2] + 176484[r12q5l3] + 111270[r12q4l4] + 4262[r11q9] + 37854[r11q8l] + 151416[r11q7l2] + 352968[r11q6l3] + 529452[r11q5l4] + 2518[r10q10] + 46252[r10q9l] + 208512[r10q8l2] + 554520[r10q7l3] + 971292[r10q6l4] + 583784[r10q5l5] + 116398[r9q9l2] + 693150[r9q8l3] + 1386300[r9q7l4] + 1940568[... |
the Young diagrams more convenient to use. (Young diagrams earned a distinct name due to their use in visualizing more complicated structures known as Young tableaux, where the boxes are filled with integers according to particular rules.) 2.8 More Numbers 221 (a) λ = (6, 4, 4, 2, 1). (b) λ = (5, 4, 3, 3, 1, 1). FIGURE ... |
, there are four different compositions of n = 3: 3, 2, 1, 1, 2, and 1, 1, 1. Let cn denote the number of compositions of n, and let cn,k denote the number of compositions of n into exactly k parts. (a) Compute the value of cn,k for each k and n with 1 ≤ k ≤ n and 1 ≤ n ≤ 5 by listing all the compositions, and then cal... |
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